QT tableview问题

Question

使用QT制作学生管理系统时，执行刷新或修改tableview后，tableview会变空白，控制台会输出

QSqlQuery::prepare: database not open
QSqlQuery::exec: database not open

但是刚进入界面时tableview有内容
贴上代码，还请各位帮忙解决

#ifndef INFOWINDOW_H
#define INFOWINDOW_H

#include <QWidget>
#include <QDebug>
#include <QString>
#include <QMessageBox>
#include "mydatabase.h"
#include "addstudent.h"
#include "deletestu.h"

namespace Ui {
class infowindow;
}

class infowindow : public QWidget
{
    Q_OBJECT

public:
    explicit infowindow(QWidget *parent = nullptr);
    ~infowindow();

private slots:
    void on_pushButton_clicked();

    void on_pushButton_3_clicked();

    void on_pushButton_2_clicked();

    void on_pushButton_4_clicked();

private:
    Ui::infowindow *ui;
    addstudent *a = new addstudent();
    deletestu *d = new deletestu();
    QSqlDatabase& db_info = Mydatabase::getDatabase();
    QSqlTableModel *m = new QSqlTableModel(this);
};

#endif // INFOWINDOW_H
//此节目头文件

#include "infowindow.h"
#include "ui_infowindow.h"

infowindow::infowindow(QWidget *parent) :
    QWidget(parent),
    ui(new Ui::infowindow)
{
    ui->setupUi(this);
    if(!db_info.open())
    {
        qDebug()<<"error";
    }
    m = new QSqlTableModel;
    m->setTable("base_infomation");
    m->select();
    ui->tableView->setModel(m);
}

infowindow::~infowindow()
{
    db_info.close();
    delete ui;
    delete a;
    delete d;
}

void infowindow::on_pushButton_clicked()
{
    a->show();
}

void infowindow::on_pushButton_3_clicked()
{
    m->select();
    ui->tableView->setModel(m);
}

void infowindow::on_pushButton_2_clicked()
{
    d->show();
}

void infowindow::on_pushButton_4_clicked()
{
    QString search_name = ui->searchline->text();
    m->setFilter(search_name);
    m->select();
    ui->tableView->setModel(m);
}
//此界面代码实现

QSqlDatabase& Mydatabase::getDatabase()
{
    if(!db.isValid())
    {
        db = QSqlDatabase::addDatabase("QMYSQL");
        db.setHostName("localhost");
        db.setPort(3306);
        db.setDatabaseName("students");
        db.setUserName("root");
        db.setPassword("PASSWORD");
    }

    return db;
}

//mydatabse类的实现

Answer 1

和db声明的位置有关吧..
类似这个博客里的https://blog.csdn.net/haigear/article/details/120416960